Puzzle for April 11, 2021 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
Once again, we thank Abby S (age 10) for sending us a challenging puzzle! Thank you, Abby!
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Hint #1
eq.4 may be expressed as: E = (A + B + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × E = 3 × ((A + B + F) ÷ 3) which becomes eq.4a) 3×E = A + B + F
Hint #2
eq.3 may be written as: (A × E) – A = A + B + F + E In the equation above, replace A + B + F with 3×E (from eq.4a): (A × E) – A = 3×E + E which becomes eq.3a) (A × E) – A = 4×E
Hint #3
In eq.3a, add A to both sides, and subtract 4×E from both sides: (A × E) – A + A – 4×E = 4×E + A – 4×E which becomes (A × E) – 4×E = A which may be written as E × (A – 4) = A Divide both sides of the above equation by (A – 4): E × (A – 4) ÷ (A – 4) = A ÷ (A – 4) which becomes eq.3b) E = A ÷ (A – 4) (assumes A ≠ 4)
Hint #4
Confirm: A ≠ 4 ... Substituting 4 for A in eq.3a would yield: (4 × E) – 4 = 4×E which would become 4×E – 4 = 4×E Subtracting 4×E from each side of the above equation would yield: 4×E – 4 – 4×E = 4×E – 4×E which would make –4 = 0 Since –4 ≠ 0 then A ≠ 4
Hint #5
Since A ≠ 0 (from eq.2), divide both sides of eq.1 by A: A × F ÷ A = (A + E) ÷ A which becomes F = (A + E) ÷ A which becomes F = 1 + (E ÷ A) Substitute (A ÷ (A – 4)) for E (from eq.3b) in the above equation: F = 1 + ((A ÷ (A – 4)) ÷ A) which becomes eq.1a) F = 1 + (1 ÷ (A – 4))
Hint #6
Check several values for A and F in eq.1a: If A = 1 then F = 1 + (1 ÷ (1 – 4)) = 1 + (1 ÷ (–3)) = 1 + (–⅓) = ⅔ If A = 2 then F = 1 + (1 ÷ (2 – 4)) = 1 + (1 ÷ (–2)) = 1 + (–½) = ½ If A = 3 then F = 1 + (1 ÷ (3 – 4)) = 1 + (1 ÷ (–1)) = 1 + (–1) = 0 If A = 5 then F = 1 + (1 ÷ (5 – 4)) = 1 + (1 ÷ 1) = 1 + 1 = 2 If A = 6 then F = 1 + (1 ÷ (6 – 4)) = 1 + (1 ÷ 2) = 1 + ½ = 1½ If A ≥ 7 then 1 < F < 2 The above equations make either: A = 3 and F = 0 or: A = 5 and F = 2
Hint #7
Check: A = 3, and F = 0 ... Substituting 3 for A, and 0 for F in eq.1 would yield: 3 × 0 = 3 + E which would become 0 = 3 + E Subtracting 3 from both sides of the above equation would yield: 0 – 3 = 3 + E – 3 which would make –3 = E Since E is non-negative, then: E ≠ –3 which means A ≠ 3 and F ≠ 0 and therefore makes: A = 5 and F = 2
Hint #8
Substitute 5 for A in eq.3b: E = 5 ÷ (5 – 4) which becomes E = 5 ÷ 1 which makes E = 5
Hint #9
Substitute 5 for E and A, and 2 for F in eq.4a: 3×5 = 5 + B + 2 which becomes 15 = 7 + B Subtract 7 from each side of the above equation: 15 – 7 = 7 + B – 7 which makes 8 = B
Solution
Substitute 5 for A and E in eq.2: (C + D) ÷ 5 = 5 – 5 which becomes (C + D) ÷ 5 = 0 which makes C + D = 0 The only non-negative numbers for C and D that satisfy the above equation are: C = 0 and D = 0 making ABCDEF = 580052